WHY PROMETHIUM HAS NO STABLE ISOTOPES
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) ( July 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks , discovered by Gell-Mann and Zweig, I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” (2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838.68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). Here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. THE UNSTABLE NUCLIDES OF PROMETHIUM Promethium (Pm) has no stable isotopes, and does not exist in nature, except in trace quantities as a product of spontaneous fission and alpha decay of Eu-151. It is a synthetic element, first produced in 1945. Thirty-eight radioisotopes have been characterized, with the most stable being Pm-145 with a half-life of 17.7 years, Pm-146 with a half-life of 5.53 years, and Pm-147 with a half-life of 2.6234 years. Comparing the structures of Promethium with 61 protons (odd number) with the structures of Tellurium with 52 protons ( even number) we see that the structutres of promethium brake the high symmetry of the structures of Tellurium.(See my STRUCTURE OF Te-120...Te-130 ). Moreover the structures of Promethium differ fundamentally from the structures of Tellurium, since here the unstable nuclides give great spins. For example in the Tellurium the odd number of extra neutrons gives S =+1/2 , while the Pm-146 gives S = -3 . For understanding why the structures of Promethium cannot give any stable nuclide I examined carefully these structures for revealing that they are due to the reduction of the number N of blank positions. For example in order to understand these difficulties we examine the unstable structure of Pm-146 of 24 extra neutrons. So let us analyze the diagram of Pm-146 along with the three cases of the top view of the first , the second, and the third horizontal plane. ' ' DIAGRAM OF Pm-146 FORMING 25 BLANK POSITIONS ABLE TO RECEIVE 14 EXTRA NEUTRONS OF POSITIVE SPINS AND 11 EXTRA NEUTRONS OF NEGATIVE SPINS Here the additional p61n61 of S=-1 is shown near the n23p23. For symmetrical arrangements the p39n39 is moved from the square to the central parallelepiped in order to fill the blank position formed by the n24p24. Here you see also the p47n47 along with the p48n48, which make two symmetrical alpha particles of opposite spins. But you cannot see the additional p49n49 the n52p52 of the third alpha particle and the n50p50 and the p51n51 of the fourth alpha particle. Also the p41, n41, p42, n42, p43, n43, p44, and n44 which form the central parallelepiped of opposite spins are not shown. In the same way the 8 deuterons of opposite spins from p13n13 to p20n20 and the 4 deuterons from p33n33 to p36 n36 are not shown. ' n40.....p40..........n ' ' n..........p38.......n38 H. Square with n' ' n31………p12.........n12.......p32.......n' ' p31........n11.........p11…… n32 Sixth h. plane' ' n..........p29.........n10.........p10…… n30 ' ' n29….....p9..........n9 …….p30.......n Fifth h. plane' ' p47.......n27.........p8..........n8.........p28........... n48' ' n45...........p27........n7.........p7........n28..........p46 Fourth h. plane ' ' n47..........p25.........n6.........p6..........n26...........p48' ' p45.......n25……….p5..........n5……….p26.........n46 Third h. plane' ' p61..............n23………p4........n4………….p24..............p39' ' n61........p23…….....n3………....p3………..n24 ...............n39 Second h. plane' ' p21.........n2………p2............n22' ' n21........p1........n1.........p22] First h. plane' ' n...............p37......n37 p37n37 with n' ' ' ' '''TOP VIEW OF THE FIRST HORIZONTAL PLANE IN WHICH ALL NUCLEONS ARE SHOWN ' HERE THE FIRST EXTRA NEUTRON (n ) OF WEAK BONDS MAKES THE TWO RADIAL BONDS WITH p22 AND p33 WHILE THE SECOND ONE MAKES THE TWO RADIAL BONDS WITH p21 AND p34. ''' (n)........p34....... n34 ' p21....... n2........ p2....... n22 ' ' n21.........p1. .......n1.......p22' ' n33.......p33..... (n)' ' TOP VIEW OF THE SECOND HORIZONTAL PLANE' Here you see the 2n of strong bonds near the p13 and p14. While the 2{n} of three bonds per neutron are near the p14 and p24 as well as near the p13 and p23 . Here you also can see the p61n61 of S=-1 and the symmetrical p39n39 of S = -1. ' n' ' n14.......p14........{n}' ' p61.......n23.......p4.........n4.........p24...........n39 ' ' n61.......p23........n3........p3.........n24............p39' ' {n}...... p13......n13 ' ' n ' TOP VIEW OF THE THIRD HORIZONTAL PLANE WITH POSITIVE SPINS ' (n.)........p58....... n50.......p51....n60 ' ' p53........n42........p16......n16......p44.........n54' ' n47........p25........n6........p6........n26.........p48' ' p45........n25........p5........n5........p26........ n46' ' n55........p41.......n15.......p15.......n43...... .p56' ' n57.......p49.......n52...... p59........(n)' Here you can see the following characteristics of the fundamental shapes formed by the nucleons of the central parallelepiped as The p5n5 and n6p6 create the small horizontal square of Mg-24 for creating the central parallelepiped of the alpha particle nuclei. The n15p15 and p16n16 create the first small horizontal rectangle. The p25n25 and p26n26 create the second small horizontal rectangle. The p41, n42, n43 and p44 make the great horizontal square of the great central parallelepiped. The p45, n46, n47 and p48 form the first great horizontal rectangle. The p49, n50, p51 and n52 form the second great horizontal rectangle. Then these great horizontal rectangles which give the structure of Tellurium with 52 protons are able to form 8 blank positions for receiving 8 vertical pn systems. Here you see the p53, n54, n55, p56 n57, p58, p59 and n60 of such vertical pn systems. In the structure of the Tellurium the n54, n55, n57 and n60 of the third plane along with the symmetrical positions of the fourth horizontal plane were able to receive 8 extra neutrons . However in the structure of Pm looking at the top view of the third horizontal plane one concludes that there exist only 4 blank positions of the third and the fourth plane for receiving 4 extra neutrons of weak binding energy. Under this condition the p61n61 of S=-1fills the blank position formed by n23p23. Moreover for symmetrical arrangements the p39n39 of S=-1 is moved from the square to the central parallelepiped in order to fill the blank positions formed by the n24p24. Thus using the diagram of Pm and the three cases of the top view like the first, the second, and the third horizontal plane in this new arrangement the number N of blank positions is given by The p37n37 gives 3n of strong bonds with negative spins The p38n38 and n40p40 give 4n of strong bonds with positive spins The first and the sixth plane give 4(n) of opposite spins. The second horizontal plane gives 2{n} + 2n with negative spins because here the p61n61 and p39n39 represent the two deuterons of S =-2. The fifth horizontal plane gives 2{n} + 4n with positive spins The third and the fourth plane give 4(n) of opposite spins That is N = 4{n} + 13n + 8(n) = 25 blank positions able to receive 14 extra neutrons of positive spins and 11 extra neutrons of negative spins ' ' WHY THE Pm-146 IS AN UNSTABLE NUCLIDE Since the Pm-146 of 24 extra neutrons has S = -3 we conclude that it is due to the summation of S = - 2 of the two new deuterons like the p61n61 and the p39n39 and the S =-1 of the extra neutrons. Thus the 25 blank positions of this structure are filled by 11 extra neutrons of positive spins and 13 extra neutrons of negative spins. In other words this number of 13 extra neutrons with negative spins is greater than the 11 blank positions of the same negative spins . So the Pm-146 needs 2 more extra neutrons of single bonds which lead to the decay. Category:Fundamental physics concepts